Jones M., Fleming S.A. Organic Chemistry

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7.8 The Unimolecular Elimination Reaction: E1 299

7.8b Selectivity in the E1 Reaction In many E1 reactions,more than one hydro-

gen can be removed from the carbocationic intermediate to give an alkene.When there

is a choice, the major product is the more substituted, more stable alkene (Chapter 3,

p. 115). Of course, statistical effects will also influence the product mix, but the over-

whelming factor is alkene stability.The more substituted alkene is favored strongly in

the E1 reaction. For example, 2-bromo-2-methylbutane reacts in a solution of ethyl

alcohol and water to give a mixture of 2-methyl-2-butene and 2-methyl-1-butene in

which the more substitued alkene predominates by a factor of 4 (Fig. 7.72).

high affinity for a proton.Good nucleophiles will be less effective in the elimination reac-

tion,which requires removal of a proton,but more effective in the S

1 reaction (Fig.7.71).

SOME SPECIFIC EXAMPLES

–

OCH

20 80

93 7

%E1

FIGURE 7.71 The elimination and

1 reactions compete. In pure

ethyl alcohol, the ratio of alkene

(2-methylpropene) to S

1 product

(tert-butyl ethyl ether) is 0.25. In the

presence of the much stronger base

sodium ethoxide, this ratio increases

to about 13. A strong base (ethoxide)

favors the elimination reaction.

THE GENERAL CASE

A SPECIFIC EXAMPLE

More substituted

(major product)

Less substituted

(minor product)

6 Hydrogens 2 Hydrogens

2-Methyl-2-butene (32%)

2-Methyl-1-butene (8%)

– H

2-Bromo-2-methylbutane

FIGURE 7.72 The more substituted

(more stable) alkene is favored in the

E1 reaction, even when statistical

factors favor the less substituted

isomer. This reaction is known as

Saytzeff elimination.

PROBLEM 7.21 Draw arrow formalisms for the reactions of Figure 7.72.

300 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM 7.22 The specific example in Figure 7.72 also produces two products of

the S

1 reaction: one C

O and the other C

O. What are these products

and how are they formed?

To see why the more substituted alkene is favored, look at the transition state

for proton removal. In the transition state the carbon–hydrogen bond is lengthened

and the double bond of the alkene is partially formed (Fig. 7.73).

Saytzeff can be transliterated from the Cyrillic as Zaitzev, Saytzev, or Saytseff.

CCH

Partially formed

double bonds

Transition states for proton removal by a

base to give the alkene and NuH

loss of

CCH

–

Nu H

FIGURE 7.73 In the transition state, a

double bond is partially formed. The

increased stability of more substituted

double bonds will be felt, and the

transition state for formation of the

more substituted alkene (loss of H

)

will be lower in energy than the

transition state for formation of the

less substituted alkene (loss of H

The factors that operate to stabilize alkenes with fully formed double bonds are

also operative in the transition states in which there are partially formed double bonds.

So, the more alkyl groups, the better, and the E1 reaction preferentially produces the

most substituted alkene possible.This phenomenon is called Saytzeff elimination,after

Alexander M. Saytzeff (1841–1910), the Russian chemist who discovered it.

PROBLEM 7.23 Predict the products of the E1/S

1 reactions of the following

molecules in water:

PROBLEM 7.24 It was noticed as early as 1862 that treatment of isobutyl halides,

, with silver acetate, Ag



OAc (



OAc is the acetate ion, a

weak nucleophile, and Ag



is a strong Lewis acid, especially toward chloride),

produced tert-butyl acetate, , and isobutene. Use the stability

order of carbocations (p. 292) to provide an explanation for this behavior. Hint:

Reason backward from the structure of the product. What ion must be present to

form this product?

(CH

)

OAc

(CH

)

CHCH

(a)

(b)

(c)

(CH

)

7.9 The Bimolecular Elimination Reaction: E2 301

7.9 The Bimolecular Elimination Reaction: E2

As with the substitution reaction, there are two common mechanisms for the elim-

ination reaction. We have just explored the first-order process, the E1 reaction.Now

let’s look at the bimolecular reaction, the E2 reaction.

7.9 a Rate L aw The S

2 reaction is also complicated by alkene formation. Just

as we saw for the S

2 reaction, the typical rate law for the bimolecular E2 reaction

shows that the alkene is formed in a second-order process that is first order in both

substrate and nucleophile:

ν  k[substrate] [nucleophile]

The mechanism involves attack by the base at a hydrogen attached to a carbon

adjacent to the carbon bearing the leaving group. Figure 7.74 shows the arrow

formalisms for the competing E2 and S

2 reactions.

Bimolecular elimination: E2

–

FIGURE 7.74 The competing

E2 and S

2 reactions.

This picture gives us an immediate clue as to how to influence the product mix-

ture from competing E2 and S

2 reactions. As in the E1–S

1 competition (p. 298),

strong Brønsted bases (high affinity for hydrogen 1s orbital) should favor the E2

reaction and strong nucleophiles (high affinity for carbon 2p orbital) should favor

the S

2 reaction (Fig. 7.75).

–

OCH

HOCH

acetone

2 Product

E2 Product

(75%)

2 Product

(25%)

CCH

FIGURE 7.75 The E2 and S

reactions compete. In the examples

shown, the relatively good

nucleophile chloride gives exclusively

the S

2 reaction, whereas the much

more basic ethoxide ion gives mostly

the E2 reaction. Bu means butyl.

7.9b Effect of Substrate Branching on the E2–S

2 Mix The rate of the

2 reaction depends strongly on the structure of the substrate. The more highly

branched the substrate, the more hindered is the obligatory approach from the rear of

the leaving group and the slower is the S

2 reaction.This effect reaches a limit with

tertiary substrates, in which the S

2 rate is effectively zero.The slower the S

2 reac-

tion, the more important is the competing E2 reaction. With a typical tertiary sub-

strate such as tert-butyl bromide, the E2 reaction is the only effective reaction under

302 CHAPTER 7 Substitution and Elimination Reactions

E2–S

2 conditions.The steric requirements for deprotonation are far less severe than

for attack at the rear of the carbon–bromine bond of tert-butyl bromide (Fig. 7.76).

–

No S

FIGURE 7.76 For tertiary substrates, the S

2 reaction does not proceed, but the E2 reaction is relatively unhindered.

Attack from the rear in S

2 fashion is blocked by the three methyl groups.The E2 reaction is relatively easy.

(49%)

(51%)

–

acetone

–

acetone

With a secondary substrate, the S

2 and E2 reactions are competitive

With a primary substrate, the S

2 reaction is often the only process

(CH

)

CH (CH

)

(CH

)

FIGURE 7.77 For secondary

substrates, the E2 and S

2 reactions

are competitive. For primary

substrates, the S

2 reaction

dominates.

7.9c Stereochemistry of the E2 Reaction When we explored the

mechanism of the S

1 and S

2 reactions, an examination of the stereochemical

outcome of the reactions was crucial to our development of their mechanisms.

Stereochemical experiments are common in organic chemistry and are one of the

reasons we spent so much time and effort earlier on stereochemical analysis. In the

E2 reaction, there are four possible extreme orientations. They can be classified by

noting the dihedral angle between the two groups that come off, the hydrogen and

the leaving group L (Fig. 7.78).

C CC

180⬚ 120⬚ 60⬚ 0⬚

Dihedral angle between C H and C L

H H

FIGURE 7.78 Four possible

arrangements for the E2 reaction.

The Newman projections emphasize

the dihedral angle between the

and bonds involved in

the reaction.

With less hindered substrates, the S

2 reaction can compete much more

effectively with the E2 reaction, and with primary substrates the S

2 reaction is

dominant (Fig. 7.77).

7.9 The Bimolecular Elimination Reaction: E2 303

As the and bonds lengthen in this reaction, the molecule flat-

tens out as the 2p orbitals that will constitute the π system of the alkene are formed.

Remember (p. 105): In the product alkene the 2p orbitals are perfectly lined up,

eclipsed, with a 0° dihedral angle (Fig. 7.79).

Transition state

Newman projection

CCCC

C C

FIGURE 7.79 The generation of a π bond as the E2

reaction occurs. In the E2 reaction, the

and bonds lengthen as the π bond between

the carbons is created. Note especially the

requirement for orbital lineup as the π bond is

generated.The dihedral angle between the two

p orbitals in the product alkene is 0°.

0⬚ syn

180⬚ anti

Dihedral

angle

= 180⬚

Dihedral

angle = 0⬚

Dihedral

angle = 60⬚

Dihedral

angle = 120⬚

60⬚ gauche

120⬚ gauche

(a) (c)

(d)(b)

FIGURE 7.80 Four views of the transition state for the E2 reaction. In examples (a) and (b), where the dihedral angle

is 0° or 180°, the π bond can be generated without an extra rotation. In examples (c) and (d), where the dihedral angle

is 60° or 120°, an extra rotation must take place in order for the π bond to be made.

Note that only the 0° (syn) and 180° (anti) arrangements shown in Figure 7.78

can lead smoothly to an alkene with perfectly overlapping 2p orbitals.The 120° and

60° forms cannot produce alkene without an additional rotation to line up the orbitals

(Fig. 7.80).

Thus, we can anticipate that the 60° and 120° gauche arrangements for

elimination should be disfavored. It is much more difficult to decide which of

the remaining two possibilities, 0° (syn) or 180° (anti) will be preferred in the E2

reaction.

304 CHAPTER 7 Substitution and Elimination Reactions

The favored backside

displacement in the

2 reaction

The disfavored

frontside displacement

in the S

2 reaction

The 180⬚ E2 reaction:

the electrons in the C

H bond displace

the leaving group from the rear

The 0⬚ E2 mimics the frontside S

2 reaction:

the electrons in the C

H bond displace

the leaving group from the front

–

FIGURE 7.81 The 0° E2 reaction

resembles a frontside S

2 reaction,

whereas the 180° E2 reaction looks

like the favored, backside S

displacement reaction.

The E2 is a kind of extended S

2 reaction. The Lewis base (Nu ) attacks the

hydrogen, and the pair of electrons in the carbon–hydrogen bond does the displac-

ing of the leaving group. If we keep this analogy in mind we can see that the 180°

elimination involves backside displacement, as does the S

2 reaction,but the 0° elim-

ination requires frontside displacement. We might therefore expect the 180° E2 to

be favored (Fig. 7.81).

2-Chloro-2,3-dimethylbutane

CCCCl

–

FIGURE 7.82 In this reaction there is

only one possible tetrasubstituted

alkene product.

Now the problem is to find an experiment that will let us see this level of detail.

As usual, we will use a stereochemical analysis, and this recquires a certain com-

plexity of substitution in the substrate molecule. For example, there is no use in

examining the stereochemistry of the E2 reaction of a molecule as simple as

2-chloro-2,3-dimethylbutane because there is only one tetrasubstituted alkene

possible (Fig. 7.82).

PROBLEM 7.25 What other alkene could be formed in the reaction of

Figure 7.82?

Molecules more highly substituted than 2-chloro-2,3-dimethylbutane can reveal

the stereochemical preference, however. For example, consider the molecules of

7.9 The Bimolecular Elimination Reaction: E2 305

Figure 7.83. In diastereomer A,syn (0°) elimination would lead to a product in which

the methyl groups are cis [this is,however, the (E) alkene], whereas anti (180°) elimina-

tion would give the compound with trans methyls [the (Z) alkene]. In B, these prefer-

ences are reversed:syn elimination would produce the (Z) alkene,and anti elimination would

give the (E) alkene. Remember: The syn form is an eclipsed, energy-maximum structure

and the anti form is a staggered,energy-minimum structure.Many tests of this kind have

been carried out,and it is certain that in acyclic molecules anti (180°) elimination is favored

over the syn process (0°) by very large amounts, typically greater than 10

(Fig. 7.83).

Tosylate, an excellent

leaving group (see Fig. 7.41)

= =

(Z)

(E)

syn

= 0⬚ anti = 180⬚

(E)(Z) syn

= 0⬚

anti

= 180⬚

O SO

OTs

WEB 3D

FIGURE 7.83 The possibilities for 0°

and 180° E2 reactions for the two

diastereomers A and B shown.

WORKED PROBLEM 7.26 In chemistry, it is customary to report one’s research

results by submitting papers to the chemical literature.As a kind of quality control,

these papers are sent to experts in the field for comment and review before publi-

cation. Imagine that you are such a referee looking over the experiment described

in Figure 7.83. Do the data absolutely require the conclusion that anti (180°) elim-

inations are preferred to syn (0°)? Obviously, we think not. What do you think?

ANSWER There is a problem. The analysis of Figure 7.83 shows a 180° elimina-

tion proceeding from a staggered, energy-minimum form. By contrast, the 0°, syn

elimination of Figure 7.83 proceeds from a fully eclipsed arrangement, which is a

transition state, not an energy minimum (p. 67; Fig. 7.22, p. 273). This compari-

son is scarcely fair! In order to make a real comparison of the syn and anti E2

processes, we will have to be more clever.

syn = 0⬚ anti = 180⬚

Eclipsed, energy maximum

form—no lifetime!

Staggered, energy

minimum form

OTs

306 CHAPTER 7 Substitution and Elimination Reactions

OTs

(98%)

–

OTs

syn, – DOTs

The dihedral angle between H and

OTs is 120⬚, far from a perfect 180⬚

anti elimination

anti, – HOTs

syn, Perfect 0⬚ angle

between D and OTs

–

WEB 3D

FIGURE 7.84 Sometimes syn elimination can be preferred.

It is important to remember that orbital overlap is vital in the E2 elimination

reaction,and a syn elimination in which the orbitals overlap well may compete effec-

tively with an anti elimination in which the perfect 180° dihedral angle cannot be

attained. For example, some polycyclic molecules give syn elimination.In the exam-

ple shown in Figure 7.84,the 2-bicyclo[2.2.1]heptyl, or 2-norbornyl, system, the ideal

anti (180°) elimination (loss of HOTs) is impossible, because only a 120° arrange-

ment of the departing groups can be achieved. By contrast, the syn elimination (loss

of DOTs) has a perfect 0° dihedral angle between the two departing groups. The

result is that elimination is very slow and the perfect syn elimination is preferred to

the 120° anti elimination (Fig. 7.84).

PROBLEM 7.27 Draw all nine possible stereoisomers of 1,2,3,4,5,6-hexachloro-

cyclohexane. First, draw the molecules flat using some schematic device to indi-

cate the stereochemistry of the chlorine atoms. Next, make three-dimensional

drawings of these molecules.

PROBLEM 7.28 One of the isomers you drew in Problem 7.27 reacts much more

slowly in the E2 reaction than all the others.Which one is it, and why is it so slow

to undergo an E2 reaction? Hint: Remember the steric requirements for the E2

reaction.

–

HOCH

(19%)

Both cis and trans

isomers are formed

CH CHCH

(81%)

FIGURE 7.85 In the E2 reaction, it is the more substituted alkene that is usually formed

preferentially. Both cis and trans isomers are formed in this example.

7.9d Selectivity in the E2 Reaction Generally, Saytzeff elimination is the

rule for the E2 elimination. The most substituted, and therefore most stable,

alkene is the major product in this regioselective reaction (Fig. 7.85). Recall that

7.9 The Bimolecular Elimination Reaction: E2 307

regiochemistry simply refers to the outcome of a reaction in which more than

one orientation of substituents is possible in the product, and in a regioselective

reaction, one product predominates.

In the transition state for alkene formation, the π bond is partially developed.

The factors that make more substituted alkenes more stable than less substituted

alkenes will operate on the partially developed alkenes in the transition states as

well. The more substituted the partially developed alkene, the more stable it is,

and this effect inevitably favors the more substituted alkene as the major product

(Fig. 7.86).

–

OCH

HOCH

Major product

–

In this transition state for the formation

of the major product, there is a partially

formed disubstituted double bond

HOCH

Minor product

–

In this transition state for the formation

of the minor product, there is a partially

formed monosubstituted double bond

–

OCH

Many factors influence the ratio of alkene products, however, and some

are quite surprising at first. The size of the base makes a small difference,

with larger bases favoring the less stable alkene. Although there is only a

very small change in the product distribution when the base is changed

from methoxide to tert-butoxide in the reaction of 2-iodobutane with

FIGURE 7.86 In the E2 reaction, the more stable transition state leads to the major product,

the more substituted alkene.

308 CHAPTER 7 Substitution and Elimination Reactions

THE GENERAL CASE

A SPECIFIC EXAMPLE

4.9

trans/cis

= 3.1

3.8

trans/cis

= 3.0

Saytzeff elimination

Hofmann elimination

dimethyl sulfoxide

(DMSO)

–

OCH

DMSO =

DMSO

–

OC(CH

)

Size makes little difference in the approach

to one of the hydrogens whose removal

leads to the less substituted alkene

There are greater steric problems in

approaching one of the hydrogens in the

interior of the molecule; thus size of the

base operates to favor the less

substituted alkene (Hofmann elimination)

FIGURE 7.87 The effect of increasing

base size is to increase the amount of

the less stable “Hofmann” product

somewhat.

alkoxide (Fig. 7.87), when the base is made truly enormous these effects can be magni-

fied. In order to form the more substituted alkene, the base must penetrate farther into

the middle of the substrate. Formation of the less stable alkene requires attack by base

farther toward the periphery of the molecule and thus has lower steric requirements.

2-Alkene

(%)

1-Alkene

(%)

19 (Saytzeff

preferred)

28 (Saytzeff

preferred)

33 (Saytzeff

preferred)

70 (Hofmann

preferred)

(cis and trans isomers)

Hofmann

Major

(98%)

Major

(69%)

Saytzeff

Minor

(31%)

–

HOCH

–

OCH

–

OCH

N(CH

)

(cis and trans isomers)

Minor

(2%)

(cis and trans isomers)

FIGURE 7.88 The regioselectivity of the E2

reaction depends on the identity of the leaving

group. Leaving groups such as fluoride,

ammonium (R



), and sulfonium (R



)

lead to predominant Hofmann elimination, in

which the less stable isomer is the major product.

Hofmann elimination

Formation of the less stable alkene is called Hofmann elimination after the man

who discovered this preference, August W. von Hofmann (1818–1892). Hofmann

elimination contrasts with the more usual Saytzeff process, in which the more sta-

ble alkene is the major product. Hofmann elimination is favored by certain leaving

groups, however. Typical examples of Hofmann leaving groups are fluoride and the

’onium ions, such as ammonium and sulfonium (Fig. 7.88).